PARALLELOGRAM
Parallelogram
Module 1: Conditions that Make a Quadrilateral a Parallelogram.
Examples are: Parallelogram, Rectangle, Square, Rhombus, and Kite.
A parallelogram is a quadrilateral with two pairs of opposite sides that are parallel.
CONDITIONS THAT MAKE A QUADRILATERAL A PARALLELOGRAM
1. A quadrilateral is a parallelogram if both pairs of opposite sides are congruent.
In the figure, the pairs of opposite sides are 𝐴𝑅̅̅̅̅ & 𝐶𝐸̅̅̅̅; 𝐴𝐶̅̅̅̅ & ̅𝑅𝐸̅̅̅. Then, measure of 𝐴𝑅̅̅̅̅ = 6.40𝑐𝑚 𝑎𝑛𝑑 measure of 𝐶𝐸̅̅̅̅ = 6.40𝑐𝑚; also measure of 𝐴𝐶̅̅̅̅ = 3.43𝑐𝑚 𝑎𝑛𝑑 measure of ̅𝑅𝐸̅̅̅ = 3.43𝑐𝑚. It shows that the pairs of opposite side of CARE are congruent. Therefore, quadrilateral CARE is a parallelogram.
2. A quadrilateral is a parallelogram if both pairs of opposite angles are congruent.
In the figure, the pairs of opposite angles are < 𝐴 𝑎𝑛𝑑 < 𝐻; < 𝑀 𝑎𝑛𝑑 < 𝑇. Then, measure of < 𝐴=120° 𝑎𝑛𝑑 measure of < 𝐻=120°; also measure of < 𝑀=60° and measure of < 𝑇=60°. It shows that the pairs of opposite angles of MATH are congruent. Therefore, quadrilateral MATH is a parallelogram.
3. A quadrilateral is a parallelogram if both pairs of consecutive angles are supplementary.
Supplementary angles are two angles having a sum of 180°. It shows that pairs of consecutive angles of MATH are supplementary. Therefore, quadrilateral MATH is a parallelogram.
4. A quadrilateral is a parallelogram if the diagonals bisect each other.
A diagonal is a line segment joining two nonconsecutive vertices of a polygon. Bisect means to divide into two equal parts. A line segment bisects another line segment if it divides the segment into two congruent parts.
5. A quadrilateral is a parallelogram if each diagonal divides a parallelogram into two congruent triangles.
Two triangles are congruent if their corresponding sides are equal in length, and their corresponding angles are equal in measure.
6. A quadrilateral is a parallelogram if one pair of opposite sides are both congruent and parallel.
In the figure, the pair of opposite side of ROSE is 𝑂𝑆 ̅̅̅̅ 𝑎𝑛𝑑 ̅𝑅𝐸̅̅̅ and they are congruent and parallel. Therefore, quadrilateral ROSE is a parallelogram.
Module 2: Using Properties to Find the Measures of Angles, Sides and Other Quantities Involving Parallelograms.
Properties of Parallelogram
1. In a parallelogram, any two opposite sides are congruent.
2. In a parallelogram, any two opposite angles are congruent.
3. In a parallelogram, any two consecutive angles are supplementary.
4. The diagonals of a parallelogram bisect each other.
5. A diagonal of a parallelogram forms two congruent triangles.
We are going to use these properties to find the measures of angles, sides, and other quantities of a parallelogram.
Examples:
1. Given ABCD as shown below. Find the measure of side 𝐴𝐷̅̅̅̅.
Since the opposite side of 𝐴𝐷̅̅̅̅ is 𝐵𝐶̅̅̅̅. And measure of 𝐵𝐶̅̅̅̅ = 6.2cm, therefore 𝑨𝑫̅̅̅̅ = 6.2cm using parallelogram property 1.
2. Below is the parallelogram ABCD. Consider each given information and answer the questions that follow.
Given: 𝐴𝐵̅̅̅ = (3x – 5)cm 𝐵𝐶̅̅̅̅ = (2y – 7)cm 𝐶𝐷̅̅̅̅ = (𝑥 + 7)𝑐𝑚 A𝐷̅̅̅̅ = (y+3)cm.
A. What is the value of x? Solution: ̅𝐴𝐵̅̅̅ ≅ 𝐶𝐷̅̅̅̅ -------- Use parallelogram property 1
3x – 5 = x + 7 (3x -x) (– 5 + 5)= (x – x) (+ 7 + 5) 2x = 12 2 2 x = 6
B. What is the value of y? Solution: 𝐴𝐷̅̅̅̅ ≅ 𝐵𝐶̅̅̅̅ -------- Use parallelogram property 1
y + 3 = 2y – 7 y – 2y + 3 – 3 = 2y -2y – 7 – 3 -y = -10 y = 10
C. How long is ≅ 𝐴𝐷̅̅̅̅? Solution: 𝐴𝐷̅̅̅̅ = (y + 3)cm 𝐴𝐷̅̅̅̅ = (10 + 3)cm 𝐴𝐷̅̅̅̅ = 13cm
Module 3: Proving Theorems on the Different Kinds of Parallelogram.
RECTANGLE
Theorem
A parallelogram is a rectangle if and only if its diagonals are congruent
| STATEMENT | REASONS |
|---|---|
| 1. ABCD is a rectangle | Given |
| 2. AD = BC | Property of a rectangle |
| AB = AB | Reflexive Property |
| 4. 4. DAB and CBA are right angles | Definition of a rectangle |
| 5. DAB = CBA | All right angles are congruent |
| 6. Triangle DAB = CBA | SAS Postulate |
| AC = BD | CPCTC |
Now that we have verified that the diagonals of a rectangle are congruent, let us try to apply this theorem.
Example: The rectangular gate has diagonal braces.
Given |GI|=47 in and |HJ|=3x+2 in. Find the value of x and |HJ|
Solution: Since diagonals of a rectangle congruent, the |HJ|=|GI|. Thus, |HJ|=47. Solving for x:
3x = 47 – 2 3x = 45 x = 15
Module 4: Proving Theorems On the Different Kinds of Parallelogram
RHOMBUS
Theorem
The diagonals of a rhombus are perpendicular to each other.
Given: Rhombus WISE with diagonals WS and EI
Prove: WS is perpendicular to EI
Theorem
The diagonals of a rhombus bisect its opposite angles.
-+Given: Rhombus MORE with diagonal EO
Prove: angles 1 = 4, angles 3 = 2
| STATEMENT | REASONS |
|---|---|
| MORE is a rhombus with diagonal EO | Given |
| MO = RE = ME = RO | All sides of a rhombus are congruent Definition of a rhombus |
| Angles M = R | Opposite angles of parallelogram are congruenty |
| Triangles MEO = ROE | SAS Postulate |
| Angles 1 = 2 and Angles 3 = 4 | CPCTC |
| Triangles MEO = ROE are isosceles triangles | Definition of isosceles triangle |
| Angles 1 = 3 and Angles 2 = 4 | Base angles of isosceles triangle are congruent |
| Angles 1 = 4 and Angles 3 = 2 | Transitive property |
Module 5: Proving Theorems On the Different Kinds of Parallelogram (Rectangle, Rhombus, Square).
SQUARE
In a square, if it is folded through a diagonal, two ≅ isosceles right triangles are formed.
Given: ABCD is a square
Prove: ΔADC ≅ ΔCBA, and ΔADC and ΔCBA are isosceles right triangles.
Module 6: Trapezoids and Kite.
Trapezoid
Trapezoid is a quadrilateral with exactly one pair of parallel sides.
The parallel sides are called bases.
The non-parallel sides are called legs.
The base angles of a trapezoid are consecutive angles whose common side is a base of the trapezoid.
Trapezoids have two pairs of base angles.
< img src="https://i.postimg.cc/htdyQS8T/16.png">The median of a trapezoid is a segment joining the midpoints of the legs of the trapezoid.
It is parallel to the bases and the length of which is equal to half the sum of the lengths of the bases.
Theorem
The median of a trapezoid is parallel to the bases and the length of which is equal to half the sum of the lengths of the bases.
In trapezoid ABCD, 𝑀𝑁 ̅̅̅̅̅̅ is the median.
Length of the median = ½ (length of the upper base + length of the lower base)
Given: Trapezoid ABCD
Median 𝑀𝑁
Prove: 𝑀𝑁̅ || 𝐴𝐵 ; 𝑀𝑁̅ || 𝐶𝐷 , and |𝑀𝑁|=|𝐷𝐶|+|𝐴𝐵| 2
PROOF:
An isosceles trapezoid is a trapezoid whose nonparallel sides are congruent.
If the legs of a trapezoid are congruent, then the trapezoid is an isosceles trapezoid.
| PROPERTY | ILLUSTRATION/EXAMPLE |
| The bases are parallel | SE//KO |
| The legs are congruent | SK = EO |
Theorem
If a trapezoid is isosceles, then each pair of base angles are congruent.
Theorem
If a quadrilateral is an isosceles trapezoid, then the diagonals are congruent.
Given: Isosceles Trapezoid ABCD
Prove: 𝐴𝐶̅̅̅̅≅𝐵𝐷
| PROPERTY | ILUSSTRATION/EXAMPLES |
| The opposite angles of an isosceles trapezoid are supplementary | ∠A and ∠C are supplementary, and ∠D and ∠B are supplementary. |
| Any lower base angle of an isosceles trapezoid is supplementary to any upper base angle. | m∠A + m∠B = 180⁰ m∠D + m∠C = 180⁰ m∠A + m∠C = 180⁰ m∠D + m∠B = 180⁰ |
KITE
A kite is a quadrilateral with two distinct pairs of consecutive sides that are congruent.
In contrast to parallelograms where opposite sides are congruent, in a kite the congruent sides are consecutive.
Theorem
If a quadrilateral is a kite, then the diagonals are perpendicular.
Given: Kite ABCD
Prove: 𝐵𝐷 ⊥ 𝐴𝐶
PROOF:
| PROPERTY | ILLUSTRATIONS/EXMPLES |
| Diagonals of a kite are perpendicular | 𝐶𝐴 ⊥ 𝐵𝐷 |
Theorem
If a quadrilateral is a kite, then it has one pair of opposite congruent angles.
Given: Kite ABCD
Prove: ∠A ≅ ∠C
PROOF:
| PROPERTIES | ILLUSTRATIONS/EXAMPLES |
| Kite has exactly one pair of opposite congruent angles | ∠BCD ≅ ∠BAD |
Theorem
If a quadrilateral is a kite, it has one diagonal that bisects the other diagonal.
Given: Kite ABCD
Prove: Diagonal 𝐵𝐷̅̅̅̅ bisects diagonal 𝐴𝐶̅̅̅̅.
| PROPERTY | ILLUSTRATIONS/EXAMPLES |
| Kite has only one diagonal that bisects the other diagonal | Diagonal 𝐾𝑇̅̅̅̅ bisects diagonal 𝐼𝐸 ̅̅̅̅. 𝑇ℎ𝑖𝑠 𝑚𝑒𝑎𝑛𝑠 𝐸𝑂 ̅̅̅̅̅≅ 𝐼𝑂̅̅̅. |
| Kite has a diagonal that bisects each of the noncongruent angles | Diagonal 𝐾𝑇̅̅̅̅ bisects ∠EKI and ∠ETI. This means ∠KTI ≅ ∠KTE and ∠IKT ≅ ∠EKT |
Module 7: Problems Involving Parallelograms, Trapezoids and Kites.
PARALLELOGRAM
Given the parallelogram with indicated measures of its sides, solve for the values of x and y. Also, solve for the perimeter.
Solution:
Since the opposite sides of a parallelogram are congruent, then
6x – 1 = 5x + 1
Therefore, x = 2
7y + 1 = 10y – 8
-3y = -9
Therefore, y = 3
Perimeter = sum of the measures of all sides.
Substitute x and y by their respective values in the expressions representing the lengths of the sides.
6x – 1 = 6(2) -1 = 12 – 1 = 11
7y + 1 = 7(3) +1 = 21 + 1 = 22
5x + 1 = 5(2) +1 = 10 + 1 = 11
10y – 8 = 10(3) -8 = 20 – 8 = 22
Thus, 11 + 11 + 22 + 22 = 66 units
TRAPEZOID
Given PQRS is a trapezoid at the right with PS = QR. If |PR| = 3x – 14cm and |SQ| = 4x – 22cm, find the length of the diagonals.
Solution:
If the given PQRS is an isosceles trapezoid, then the diagonals are congruent.
PR = SQ then |PR| = |SQ|
3x – 14 = 4x – 22
3x – 4x = -22 – 14
-x = -8
x = 8
Substitute x by its value in both equations:
KITE
SIMILARITIES
MODULE 11 (Condition for Proving Triangles Similar)
Two triangles are similar if and only if their corresponding
angles are congruent and their corresponding sides are proportional.
Angle-Angle (AA)
similarity postulate- if two angles of one triangle
are congruent to two angles of another triangle, then the triangles are similar.
Condition: ∠L ≅ ∠𝐷, ∠V ≅ ∠𝐸, ∠O ≅ ∠𝐸 Conclusion: ∆LVO~∆DEF
| Statements | Reasons |
| 1. ∠L ≅ ∠𝐷 ∠V ≅ ∠𝐸 | GIVEN |
| 2. 𝑚∠L = m∠D 𝑚∠V = m∠F | Definition of congruent angles |
| 3. 𝑚∠L + 𝑚∠V = m∠𝐷 + 𝑚∠E | APE |
| 4. 𝑚∠L + 𝑚∠V + 𝑚∠O = 180 degrees m∠𝐷 + 𝑚∠E + 𝑚∠F = 180 degrees | In any triangle, the sum of the measures of its angle is 180 degrees . |
| 5. 𝑚∠L + 𝑚∠V + 𝑚∠O = m∠𝐷 + 𝑚∠E + 𝑚∠F | Transitive property |
| 6. 𝑚∠O = 𝑚∠F | Subtraction Property of equality |
| 7. ∠O ≅ ∠F | Definition of congruent angles |
| 8. ∆LVO~∆DEF | AAA Similarity Postulate |
Example: Determine if the two triangles are similar.
Since m∠M = m∠Q = 100 and m∠O = m∠R = 20 ,
∠𝑀 ≅ ∠𝑄 and ∠𝑂 ≅ ∠𝑅.
By the AA Similarity Postulate, ∆MNO ~ ∆QPR.
SAS Similarity Theorem- states that if two sides of one triangle are
proportional to two sides of another triangle,
and the included angle between these sides is congruent, then the triangles are similar.
Given: LM/TU = LN/TV , ∠L ≅ ∠T Prove: ∆LMN~∆TUV
SSS Similarity Theorem
- if the three sides of one triangle are proportional to the
three sides of another triangle, then the triangles are similar.
Example 1: Which of the following triangles are similar?
Solutions:
Compare the sides of three triangles by checking if the sides are in proportion.
| ∆AMP and ∆UTV | ∆AMP and ∆XWY | ∆UTV 𝐚𝐧𝐝 ∆XWY |
AM/UT = MP/TV = AP/UV 6/4 ≠ 4/3 ≠ 2/2 ce the lengths of the sides are not proportional, ∆𝐴𝐵𝐶 and ∆𝑅𝑆𝑇 are not similar. |
AM/XW = MP/WY = AP/XY 6/3 = 4/2 = 2/1 Because the lengths of the sides are proportional, ∆AMP and ∆XWY are similar or ∆AMP ~ ∆XWY by SSS Similarity theorem. |
UT/XW = TV/WY = UV/XY 4/3 = 3/2 = 2/1 Since the lengths of the sides are not proportional, ∆AMP and ∆UTV are not similar |
MODULE 12 ( Right Triangle Similarity Theorems)
Right triangle
- is a triangle that has one angle measuring exactly 90 degrees.
Right Triangle Similarity Theorem
- states that if an altitude is drawn from the vertex
of the right angle to the hypotenuse of a right triangle,
all triangles formed are similar to each other.
- It also follows that these similar triangles have proportional
corresponding sides whose lengths can be found by solving
for the geometric means of the said sides.
Given is a right triangle Δ𝐴𝐵𝐶 with ∠𝐵𝐴𝐶 as its right angle,
and hypotenuse 𝐵𝐶̅. 𝐴𝐷̅ is the altitude to the hypotenuse of Δ𝐴𝐵𝐶,
the following statements are true:
ΔROS ∼ ΔERS ∼ ΔEOR
|AE|/|OE| = |SE|/|RS|
|RO|/|OE| = |OS|/|RO|
|RS|/|SE| = |OS|/|RS|
MODULE 13 (Proving Conditions for Special Right Triangles)
A 30-60-90 triangle is characterized by one 90-degree angle,
one 30-degree angle, and one 60-degree angle.
To prove the unique side length ratios of this triangle,
let us consider an arbitrary 30-60-90 triangle with
sides denoted as a, b, and c (hypotenuse).
To start with, lets’ try to prove the 30°-60°-90° right triangle theorem.
In the 30°-60°-90° triangle,
the length of the hypotenuse is twice the length of the shorter leg.
The length of the longer leg is √3 times the length of the shorter leg.
length of the hypotenuse = 2(length of the shorter leg)
length of the 𝑙𝑜𝑛𝑔𝑒𝑟 𝑙𝑒𝑔 = √3 (length of the 𝑠ℎ𝑜𝑟𝑡𝑒𝑟 𝑙𝑒𝑔)
Proof:
For 30° − 60° − 90°, ∆ABC in equilateral ∆ACD ,
AD is the perpendicular bisector of AC.
Thus, |BD| = ½ |BC| = ½ |BA|, or |BA| = 2|BC| = 2s.
|BC|^2 + |AD|^2 =|BA|^2 Use Pythagorean Theorem
s^2 + |AD|^2 = (2s)^2 Express |AB| in terms of s, thus |AB|= 2s
|AD|^2 = 4s^2 – s^2 Subtract s2 from each side
|AD|^2 = 3s^2 Simplify
|YW| = s√3 Find the square root of each side
MODULE 14 (Applying Triangle Similarity Theorems)
Triangle similarity theorems are based on the concept that if two triangles have equal corresponding angles,
then their sides are proportional. The following are illustrations of the similarity theorems that we have discussed in the previous modules.
We can use these theorems to show similarity between triangles.
AA Similarity Corollary
States that if two angles of one triangle are congruent to two angles of another triangle,
then the triangles are similar.
SSS Similarity Theorem
If the lengths of the corresponding sides of two triangles are proportional,
then the triangles are similar.
SAS Similarity Theorem
States that if two sides of one triangle are proportional to two sides of another triangle,
and the included angles are congruent, then the triangles are similar.
Triangle Proportionality Theorem
fundamental in the field of mathematics because they allow us to solve a wide range of geometric problems.
MODULE 15 (The Pythagorean Theorem)
The Pythagorean Theorem is named after the Greek mathematician Pythagoras,
who is widely regarded as one of the most influential figures in the history of mathematics.
It states that in a right-angled triangle, the square of the length of the hypotenuse is
equal to the sum of the squares of the lengths of the other two sides.
It can be expressed as a^2 + b^2 = c^2, where c represents the length of the hypotenuse,
and a and b represent the lengths of the two other sides.
EXAMPLES OF THE PYTHAGOREAN THEOREM
Use the Pythagorean Theorem to determine the value of 𝑥
in the figure at the right.
Step 1
Identify the legs and the hypotenuse of the right triangle.
The legs have lengths 8 cm and 10 cm and 𝑥 is the length of the hypotenuse,the side opposite the right angle.
Step 2
Substitute the variables by the given values into the formula
(remember ‘c’ is the length of the hypotenuse).
𝒂^𝟐 + 𝒃^𝟐 = 𝒄^𝟐
3^𝟐 + 4^𝟐 = 𝒙^2
Step 3
Solve for the unknown.
3^𝟐 + 4^𝟐 = 𝒙^𝟐
9 + 16 = 𝒙^𝟐
25 = 𝒙^𝟐
√25 = 𝒙
5 cm = x
Therefore, the length of the hypotenuse of the right triangle is 5 cm.
Use the Pythagorean Theorem to determine the value of x
Step 1
Identify the legs and the hypotenuse of the right triangle.
The legs have lengths 3 and x.
The length of the hypotenuse is 5.
Step 2
Substitute the variables by the given values into the formula
(remember ‘c’ is the length of the hypotenuse)
𝒂^𝟐 + 𝒃^𝟐 = 𝒄^𝟐
𝒙^𝟐 + 3^𝟐 = 5^2
Step 3
Solve for the unknown.
𝒙^𝟐 + 3^𝟐 = 5^𝟐
𝒙^𝟐 + 9 = 25
𝒙^𝟐 = 25 − 9
𝒙^𝟐 = 16
𝒙 = √16
x = 4
Therefore, the length of the other leg of the right triangle is 4 cm.
Example:
Give the theorem that supports the similarity of the given pairs of triangles.
Then, find the length of the indicated side.
We can say that ΔEBA ∼ ΔECD by SAS.
|ED|/|EC| = |EA|/|EB| Write the observable proportion and solve for the unknown
|EA|/4 = 5/10 Use the known values
|EA| = 8 ∙ 3/12 Solve for |EA| and simplify
|EA| = 2
MODULE 16 (Solving Problems in Triangle Similarity and Right Triangles)
Triangle similarity
refers to the concept that two triangles are considered similar
if their corresponding angles are congruent and their corresponding sides are proportional.
The concept of triangle similarity is crucial in various real-world applications.
For example, in architecture and construction,
it is used to determine the height of a tall building or the length of a
bridge by using similar triangles in conjunction with basic measurements and angles.
Another application is in map scaling,
where we can use similar triangles to calculate the actual distance between any two points on a map.
In triangle ABC
𝐴′𝐶′ is parallel to 𝐴𝐶.
Find the value of y or |BC’| and the value of x or |A’A|.
Notice:
that the given problem involves similar triangles.
We can apply the concepts that you’ve learned about the theorems
and postulates in similar triangles.
Solution:
𝐵𝐴̅ is a transversal that intersects the two parallel lines 𝐴’𝐶’ and 𝐴𝐶̅,
hence the corresponding ∠BA’C’ and ∠BAC are congruent.
̅𝐵𝐶̅ is also a transversal to the two parallel lines A’C’,
and 𝐴𝐶⃡ and therefore ∠BC’A’ and ∠BCA are congruent.
These two triangles have two congruent angles and are therefore similar
and the lengths of their sides are proportional.
Let us separate the two triangles as shown below.
We now use the proportionality of the lengths of the sides
to write equations to solve for the values of x and y.
An equation for x may be written as follows.
(20 + 𝑥)/20 = 12/6
Solve the equation for x.
120 + 6x = 240
x=20
An equation for y may be written as follows.
12/6 = (𝑦 + 5)/y
Solve the equation for y.
12𝑦 = 6𝑦 + 30
6𝑦 = 30
y = 5
PROPORTION
Module 9 ( Describe a Proportion)
Ratio
- is a comparison of two or more quantities, and it is often represented in the form of a fraction. For ex.Suppose we have a class consisting of 20 boys and 30 girls.
20 : 30 or 20/30 - we can be simplified to ⅔
Find the ratio of a to b, where a=2 and b=6
2 : 6 or 2/6 - we can be simplified to ⅓
Proportion
- is a relationship between two or more quantities that are equivalent.It is a fundamental concept that plays a significant role in various mathematical operations and real-life applications.
a:b = c:d or a/b = c/d
where b is not equal to 0, d is not equal to zero Each quantity in a proportion is called term of the proportion.
Thus, in the proportion 𝑎: 𝑏 = 𝑐: 𝑑, a, b, c, and d are the terms. b and c are called means while a and d are called extremes.
4 : 3 = 9 : 12
3 and 9 are means
4 and 12 are extremes
Examples:
Is 8:100=4:50 a proportion?
Method 1: By simplifying the ratios
8/100 = 4/50 — 2/25 = 2/25
Each of these ratios (fractions) in simplest form is 2/25 which indicates that the two ratios form a proportion.
Method 2 : By multiplying the means and multiplying the extremes
8:100=4:50 8(50) = 100(4)
400=400 (TRUE)
Determine if 5/300 = 1/60
Method 1 : By simplifying the ratios
5/300 = 1/60 — 1/60 = 1/60
Each of the ratios in simplest form is 1/60. Therefore, the given is a proportion.
Method 2 : By multiplying the means and multiplying the extremes
5/300 = 1/60 5(60) = 300(1)
300=300 (TRUE)
Properties of Proportion
The following properties show different ways of rewriting proportions that do not alter the meaning of their values.
| Cross Multiplication Property / MeansExtremes Property | If 𝑎/𝑏 = 𝑐/𝑑 then 𝑎𝑑 = 𝑐𝑏; 𝑏 ≠ 0; 𝑑 ≠ 0 |
| Alternation Property | If 𝑎/𝑏 = 𝑐/𝑑 then 𝑎/𝑐 = 𝑏/𝑑 ; 𝑏 ≠ 0; 𝑐 ≠ 0; 𝑑 ≠ 0 |
| Inverse Property / Reciprocal Property | If 𝑎/𝑏 = 𝑐/𝑑 then 𝑏/𝑎 = 𝑑/𝑐 ; 𝑎 ≠ 0; 𝑏 ≠ 0; 𝑐 ≠ 0; 𝑑 ≠ 0 |
| Addition Property | If 𝑎/𝑏 = 𝑐/𝑑 then 𝑎+𝑏/𝑏 = 𝑐+𝑑/𝑑 ; b≠ 0; 𝑑 ≠ 0 |
| Subtraction Property | If 𝑎/𝑏 = 𝑐/𝑑 then𝑎−𝑏/𝑏 = 𝑐−𝑑/𝑑 ; 𝑏 ≠ 0; 𝑑 ≠ 0 |
| Sum Property of the Original Proportion | If 𝑎/𝑏 = 𝑐/𝑑 , then 𝑎/𝑏 = 𝑐/𝑑 = 𝑎+𝑐/𝑏+𝑑 ; 𝑏 ≠ 0; 𝑑 ≠ 0 |
MODULE 10 ( Applying the Fundamental Theorems of Proportionality to Solve Problems Involving Proportion )
Proportionality
- refers to the concept of comparing different quantities and their relationships.
Lesson 1: Basic Proportionality Theorem
Basic Proportionality Theorem - was introduced by a famous Greek Mathematician, Thales,
hence it is also called Thales Theorem. According to him, for any two equiangular triangles,
the ratio of any two corresponding sides is always the same.
APPLICATION OF FUNDAMENTAL THEOREMS OF PROPORTIONALITY
In geometry, we used proportion to compare lengths of segments.
To solve for unknown length, we often used the properties of proportion.
Here are some examples on how to apply the fundamental theorems of
proportionality to solve problems involving proportions.
1: Use the proportion 𝑚/𝑛 = 4/7 to complete each proportion.
a. 𝑛/7 = ___ b. 4/𝑚 = ___ c. 𝑛/𝑚 = ___ d. 𝑛+𝑚/𝑚 = ___
Solution
a. 𝑛/7 = 𝑚/4 b. 4/𝑚 = 7/𝑛 c. 𝑛/𝑚 = 7/4 d. 𝑛+𝑚/𝑚 = 11/4
Find the value of x in the following proportions.
a. 9/𝑥 = 15/20 b. x :6 = 15 : 18 c. 𝑥+3/4 = 9/2 d. 𝑥+2/3 = 4𝑥/6
a. 9/𝑥 = 15/20 → 15 ∙ x = 9 ∙ 20 → 15x = 180 → ( 1/15)15x = ( 1/15)180 → x = 12 b. x :6 = 15 : 18 → 6 ∙ 15 = 18 ∙ x → 90 = 18x → ( 1/18)90 = ( 1/18)18x → 5 = x or x = 5 c. 𝑥+3/4 = 9/2 → 4 ∙ 9 = 2 (x + 3) → 36 = 2x + 6 → 36 – (6) = 2x + 6 – (6) → 30 = 2x → ( 1/2 ) 30 = ( 1/2 )(2x) → 15 = x or x = 15 d. 𝑥+2/3 = 4𝑥/6 → 3 ∙ 4x = 6(x + 2) →12x = 6x + 12 → 12x – 6x = 6x - 6x + 12 → 6x = 12 → ( 1/6 )6x = ( 1/6 )12 → x = 2 Solution